HDU2602Bone Collector 0-1背包问题

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#hdu2602 #acm #0-1背包问题 #贪心算法

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本文介绍了一道经典的0-1背包问题，通过给定不同骨头的价值和体积，求解骨收集者能获得的最大总价值。文章提供了完整的C++代码实现，并采用输入文件的方式进行数据读取。

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题目要求：

Problem Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

Output

One integer per line representing the maximum of the total value (this number will be less than 2³¹).

解题思路：

简单的0-1背包问题，这里不再赘述。

代码如下：

# include <iostream>
# include <algorithm>
using namespace std;

struct node
{
	int value;
	int volume;
}bone[1002];

int value[1002];

int main()
{
	freopen("input.txt","r",stdin);
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,v;
		scanf("%d%d",&n,&v);

		int i,j;
		for(i=0;i<n;i++)
			scanf("%d",&bone[i].value);
		for(i=0;i<n;i++)
			scanf("%d",&bone[i].volume);
		
		memset(value,0,sizeof(value));
		for(i=0;i<n;i++)
		{
			for(j=v;j>=bone[i].volume;j--)
			{
				value[j]=max(value[j],value[j-bone[i].volume]+bone[i].value);
			}
		}
		printf("%d\n",value[v]);

	}
	return 0;
}